Equivalent flat topped waveform

Approximation? By what criterion? What do you mean by "equivalent"?

John

The total charge per pulse and mean current will be the same. But that may not be "equivalent", depending on what you mean by "equivalent". For example, if you care about power dissipated in a resistive load then you'll dissipate less in the flat case.

Phil.

What's important, average current or average voltage or average power?

If average power, you probably want to use a RMS average.

The wonderful thing about averages is that there are so many to choose from!

Tim.

As the others said, it depends what you're doing with it. If it's average current, then the approximation is exact. If it's power in a resistor carrying that current, it's higher. If you used the average current, 15 A, in the formula I^2R, you'd get 225 R, but that would be low. The correct value would be 233.33 R*, or 3.8% higher.

-- John

• 1/(t1-t0) * int ( (10 + 10(t-t0)/(t1-t0))^2 dt over t=t0 to t1) =

Since you failed to say anything about the approximation (mathematically or with electronics), how can we even guess on its net worth for your application (which we also know nothing about)?

Thank you all for responding,

I think I need to try and explain myself a bit better, so hear it goes...

I have come across an approximation that I wanted to try and prove mathematically. The approximation is to convert trapezoidal pulse (with say high value =3D Xhv and low value =3D Xlv) to an equivalent rectangular pulse (with peak =3D Ypk). Ignore what quantity it represents, its an area under curve thing. The area should be the same in each pulse, just the shape different.

What i have for an approximation follows =3D> for the trapezoidal pulse =3D> [(Xhv*Xhv)+(Xlv*Xlv)]/2=3DYpk and am not sure if this stacks up.

thanks,

reggie

For equal area under the curve, you want Ypk = (Xhv+Xlv)/2.

That doesn't make any sense.

But this similar expression could make sense:

Ypk*Ypk = [ (Xhv*Xhv) + (Xlv*Xlv) ] / 2

This gives a pulse with the same root-mean-squared value. This would make sense if you were measuring voltage or current, and wanted equal power dissipation in a resistive load.

Phil.

Thanks Phill for your reply,

I think you have got it correct, it should be Ypk*Ypk.

how would you arrive mathematically at this equation Ypk*Ypk = [ (Xhv*Xhv) + (Xlv*Xlv) ] / 2 for a rectangular waveform of peak =Ypk*Ypk?

What is confusing me is the dividing by 2, let me explain:

I understand for a RMS value you square each of the points and divide by the number of intervals, and then take the squroot of the whole thing. is this way of finding the RMS called the mid ordinate rule? Is it an accurate approximation in this case or does it depend on the time between intervals?

I have seen in one book using mid ordinate rule dividing between number of samples taken (in this case 2) but in other books dividing by (time1 at point Xhv - time2 at point Xlv) eg (Xhv^2+Xlv^2)/(t1- t2). What is the difference?

I hope I haven't confused you,

Thanks,

Reggie.

These amount to the same thing in your case because your two pulses have the same length, so (t1-t2) cancels out.

Phil